Let $f:A \rightarrow B$ be a weak equivalence of simplicial commutative rings. There is a Quillen pair $(-\otimes_{A}B, f_{\ast})$ which is an equivalence. In this situation, $(-\otimes_{A}B, f_{\ast})$ is an equivalence precisely if the counit map

$$\eta_{M} : M \rightarrow f_{\ast}(M \otimes_{A}B)$$

is a weak equivalence for all cofibrant $A$-modules $M$. I am having trouble seeing why this map is a weak equivalence. My guess is use the sequence of $A$-modules $\ker(f) \rightarrow A \rightarrow B$ to compute the homotopy groups of the sequence $\ker(\eta_{M}) \rightarrow M \rightarrow f_{\ast}(M\otimes_{A}B)$.

Is this the right idea, or is there some easier way to demonstrate the pair is a Quillen equivalence?

Let $f:A \rightarrow B$ be a weak equivalence of simplicial commutative rings. There is a Quillen pair $(-\otimes_{A}B, f_{\ast})$ which is an equivalence. In this situation, $(-\otimes_{A}B, f_{\ast})$ is an equivalence precisely if the unit map

$$\eta_{M} : M \rightarrow f_{\ast}(M \otimes_{A}B)$$

is a weak equivalence for all cofibrant $A$-modules $M$. I am having trouble seeing why this map is a weak equivalence. My guess is use the sequence of $A$-modules $\ker(f) \rightarrow A \rightarrow B$ to compute the homotopy groups of the sequence $\ker(\eta_{M}) \rightarrow M \rightarrow f_{\ast}(M\otimes_{A}B)$.

Is this the right idea, or is there some easier way to demonstrate the pair is a Quillen equivalence?